Rs Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 6 Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Algebraic Expressions are extremely popular among Class 7 students for Maths Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 99:

(i)
5x + 7x + (-6x)
= 5x + 7x -6x
= 6x

(ii)

(iii)
5a2b +( −8a2b)  + 7a2b
= 5a2b − 8a2b + 7a2b
4a2b

(iv)

(v)
Collecting like terms and adding them:

x − 3y + 4z + y − 2x − 8z + 5x − 2y − 3z
= x- 2x + 5x - 3y + y - 2y + 4z - 8z - 3z
= 4x -4y -7z

(vi) Collecting like terms and adding them:
2x2 − 3y2 + 5x2 + 6y2 + (− 3x2 − 4y2)

(vii) Collecting like terms and adding them:
5x − 2x2 − 8 +  8x2 − 7x − 9 +  3 + 7x2 − 2x

(viii) Collecting like terms and adding them:

(ix) Collecting like terms and adding them:

(x) Collecting like terms and adding them:
x+117y+94xy,32x53y9

#### Page No 99:

(i) 7xy- (-8xy)
= 7xy+ 8xy
= 15xy

(ii)  - 3x2 - x2
= -4x2

(iii)  (4y - 5x) - (x- y)
= 4y - 5x - x + y
= 5y - 6x

(iv)
(a2 + b2 + 2ab) - (a2 + b2 2ab)
=     (Collecting like terms and adding them)
= 4ab

(v)
(2x2 − 3y2 + 6xy) -  (x2y2)
(Collecting like terms and adding them)

(vi)  (2z -x -3y) - (x - y +3z)
= 2z -3z -x -x -3y +y       (Collecting like terms and adding them)
= -z -2x - 2y

#### Page No 99:

(a + 3b − 4c) + (4ab + 9c) + (−2b + 3ca)
= a + 4a - a + 3b -b -2b -4c +9c + 3c
= 4a + 8c

Now,  (4a + 8c ) - (2a 3b + 4c)
= 4a - 2a + 3b + 8c - 4c
= 2a + 3b + 4c

#### Page No 99:

(8m − 7n + 6p2) + (−3m − 4np2)

(2m + 4n − 3p2) + (− mn − p2).

#### Page No 99:

(8a − 6a2 + 9)+  (−10a − 8 + 8a2)

Collecting like terms and adding them:

#### Page No 100:

Collecting like terms and adding them:

(i)  5x + 7x - 9y -y
= 12x -10y

(ii)

(iii)  7 + 7 - 2x -x - 5x + 5y + y - 3y
= 14 - 8x -3y

(iv)

3a2 × 8a4

#### Page No 102:

−6x3 × 5x2
$=\left(-6×5\right)×\left({\mathrm{x}}^{3}×{\mathrm{x}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-30\right)×\left({\mathrm{x}}^{\left(3+2\right)}\right)\phantom{\rule{0ex}{0ex}}=-30{\mathrm{x}}^{5}\phantom{\rule{0ex}{0ex}}$

#### Page No 102:

(−4ab) × (−3a2bc)

#### Page No 102:

(2a2b3) × (−3a3b)
$=\left(2×\left(-3\right)\right)×\left({\mathrm{a}}^{2}×{\mathrm{a}}^{3}×{\mathrm{b}}^{3}×\mathrm{b}\right)\phantom{\rule{0ex}{0ex}}=\left(-6\right)×\left({\mathrm{a}}^{\left(2+3\right)}×{\mathrm{b}}^{\left(3+1\right)}\right)\phantom{\rule{0ex}{0ex}}=-6{\mathrm{a}}^{5}{\mathrm{b}}^{4}$

#### Page No 102:

$=\left(\frac{2}{3}×\frac{3}{5}\right)×\left({\mathrm{x}}^{2}×\mathrm{x}×\mathrm{y}×{\mathrm{y}}^{2\right)}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{5}×{\mathrm{x}}^{\left(2+1\right)}×{\mathrm{y}}^{\left(1+2\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{5}{\mathrm{x}}^{3}{\mathrm{y}}^{3}$

#### Page No 102:

$=\left(\frac{-3}{4}×\frac{-2}{3}\right)×\left(\mathrm{a}×{\mathrm{a}}^{2}×{\mathrm{b}}^{3}×{\mathrm{b}}^{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×{\mathrm{a}}^{\left(1+2\right)}×{\mathrm{b}}^{\left(3+4\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\mathrm{a}}^{3}{\mathrm{b}}^{7}$

#### Page No 102:

$=\left(\frac{-1}{27}×\frac{-9}{2}\right)×\left({\mathrm{a}}^{2}×{\mathrm{a}}^{3}×{\mathrm{b}}^{2}×\mathrm{b}×{\mathrm{c}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{6}×{\mathrm{a}}^{\left(2+3\right)}×{\mathrm{b}}^{\left(2+1\right)}×{\mathrm{c}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}{\mathrm{a}}^{5}{\mathrm{b}}^{3}{\mathrm{c}}^{2}$

#### Page No 102:

$=\left(\frac{-13}{5}×\frac{7}{3}\right)×\left(\mathrm{a}×{\mathrm{a}}^{2}×{\mathrm{b}}^{2}×\mathrm{b}×\mathrm{c}×{\mathrm{c}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{-91}{15}{\mathrm{a}}^{\left(1+2\right)}×{\mathrm{b}}^{\left(2+1\right)}×{\mathrm{c}}^{\left(1+2\right)}\phantom{\rule{0ex}{0ex}}=\frac{-91}{15}{\mathrm{a}}^{3}{\mathrm{b}}^{3}{\mathrm{c}}^{3}$

#### Page No 102:

$=\left(-\frac{18}{5}×\frac{-25}{6}\right)×\left({\mathrm{x}}^{2}×\mathrm{x}×\mathrm{z}×{\mathrm{z}}^{2}×\mathrm{y}\right)\phantom{\rule{0ex}{0ex}}=15×{\mathrm{x}}^{\left(2+1\right)}×\mathrm{y}×{\mathrm{z}}^{\left(1+2\right)}\phantom{\rule{0ex}{0ex}}=15{\mathrm{x}}^{3}{\mathrm{yz}}^{3}$

#### Page No 102:

$=\left(\frac{-3}{14}×\frac{7}{6}\right)×\left(\mathrm{x}×{\mathrm{x}}^{3}×{\mathrm{y}}^{4}×\mathrm{y}\right)\phantom{\rule{0ex}{0ex}}=\frac{-1}{4}{\mathrm{x}}^{\left(1+3\right)}×{\mathrm{y}}^{\left(4+1\right)}\phantom{\rule{0ex}{0ex}}=\frac{-1}{4}{\mathrm{x}}^{4}{\mathrm{y}}^{5}$

#### Page No 102:

$=\left(\frac{-7}{5}×\frac{3}{2}×\frac{-6}{5}\right)×\left({\mathrm{x}}^{2}×\mathrm{x}×{\mathrm{x}}^{3}×\mathrm{y}×{\mathrm{y}}^{2}×{\mathrm{y}}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{63}{25}×{\mathrm{x}}^{\left(2+1+3\right)}×{\mathrm{y}}^{\left(1+2+3\right)}\phantom{\rule{0ex}{0ex}}=\frac{63}{25}{\mathrm{x}}^{6}{\mathrm{y}}^{6}$

#### Page No 102:

$=\left(2×\left(-5\right)×\left(-6\right)\right)×\left({\mathrm{a}}^{2}×\mathrm{a}×\mathrm{b}×{\mathrm{b}}^{2}×\mathrm{b}×\mathrm{c}×{\mathrm{c}}^{2}\right)\phantom{\rule{0ex}{0ex}}=60×{\mathrm{a}}^{\left(2+1\right)}×{\mathrm{b}}^{\left(1+2+1\right)}×{\mathrm{c}}^{\left(1+2\right)}\phantom{\rule{0ex}{0ex}}=60{\mathrm{a}}^{3}{\mathrm{b}}^{4}{\mathrm{c}}^{3}$

#### Page No 102:

$=\left(-4×\left(-6\right)×\left(-3\right)\right)×\left({\mathrm{x}}^{2}×\mathrm{x}×{\mathrm{y}}^{2}×\mathrm{y}\right)\phantom{\rule{0ex}{0ex}}=-72×{\mathrm{x}}^{\left(2+1\right)}×{\mathrm{y}}^{\left(2+1\right)}\phantom{\rule{0ex}{0ex}}=-72{\mathrm{x}}^{3}{\mathrm{y}}^{3}$

#### Page No 102:

$=\left(\frac{-3}{5}×\frac{15}{7}×\frac{7}{9}\right)×\left({\mathrm{s}}^{2}×\mathrm{s}×\mathrm{s}×\mathrm{t}×{\mathrm{t}}^{2}×\mathrm{u}×{\mathrm{u}}^{2}\right)\phantom{\rule{0ex}{0ex}}=-1×{\mathrm{s}}^{\left(2+1+1\right)}×{\mathrm{t}}^{\left(1+2\right)}×{\mathrm{u}}^{\left(1+2\right)}\phantom{\rule{0ex}{0ex}}=-{\mathrm{s}}^{4}{\mathrm{t}}^{3}{\mathrm{u}}^{3}$

#### Page No 102:

$=\left(\frac{-2}{7}×\frac{-14}{5}×\frac{-3}{4}\right)×\left({\mathrm{u}}^{4}×\mathrm{u}×{\mathrm{u}}^{2}×\mathrm{v}×{\mathrm{v}}^{3}×{\mathrm{v}}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}×{\mathrm{u}}^{\left(4+1+2\right)}×{\mathrm{v}}^{\left(1+3+3\right)}\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}{\mathrm{u}}^{7}{\mathrm{v}}^{7}$

#### Page No 102:

$=\left(-3×-1×-1\right)×\left(\mathrm{a}×{\mathrm{a}}^{2}×\mathrm{a}×{\mathrm{b}}^{2}×{\mathrm{b}}^{2}×\mathrm{b}×\mathrm{c}×{\mathrm{c}}^{3}×\mathrm{c}\phantom{\rule{0ex}{0ex}}=-3×{\mathrm{a}}^{\left(1+2+1\right)}×{\mathrm{b}}^{\left(2+2+1\right)}×{\mathrm{c}}^{\left(1+4+1\right)}\phantom{\rule{0ex}{0ex}}=-3{\mathrm{a}}^{4}{\mathrm{b}}^{5}{\mathrm{c}}^{5}$

#### Page No 102:

$=\left(\frac{4}{3}×\frac{1}{3}×\left(-6\right)\right)×\left({\mathrm{x}}^{2}×\mathrm{x}×\mathrm{x}×\mathrm{y}×{\mathrm{y}}^{2}×\mathrm{y}×\mathrm{z}×\mathrm{z}×{\mathrm{z}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{-8}{3}×{\mathrm{x}}^{\left(2+1+1\right)}×{\mathrm{y}}^{\left(1+2+1\right)}×{\mathrm{z}}^{\left(1+1+2\right)}\phantom{\rule{0ex}{0ex}}=\frac{-8}{3}{\mathrm{x}}^{4}{\mathrm{y}}^{4}{\mathrm{z}}^{4}$

#### Page No 102:

$\frac{-2}{3}{\mathrm{a}}^{2}\mathrm{b}×\frac{6}{5}{\mathrm{a}}^{3}{\mathrm{b}}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{-2}{3}×\frac{6}{5}\right)×\left({\mathrm{a}}^{2}×{\mathrm{a}}^{3}×\mathrm{b}×{\mathrm{b}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{-4}{5}×{\mathrm{a}}^{\left(2+3\right)}×{\mathrm{b}}^{\left(1+2\right)}\phantom{\rule{0ex}{0ex}}=\frac{-4}{5}{\mathrm{a}}^{5}{\mathrm{b}}^{3}\phantom{\rule{0ex}{0ex}}$

When a =2 and b =3, we get:

L.H.S. = R.H.S.

Hence, the result is verified.

#### Page No 106:

By column method:

#### Page No 106:

By column method:

#### Page No 106:

By column method:

#### Page No 106:

By column method:

#### Page No 106:

By column method:

#### Page No 106:

By column method:

#### Page No 106:

By column method:

#### Page No 106:

(3x +4)(2x -3)

∴ (3x + 4)(2x − 3) + (5x − 4)(x + 2)

#### Page No 106:

(5x-3)(x+4)

(2x +5)(3x-4)

∴ (5x − 3)(x + 4) − (2x + 5)(3x − 4)

#### Page No 106:

(2x +5y)(3x+4y)

∴ (2x + 5y)(3x + 4y) − (7x + 3y)(2x − y)

#### Page No 106:

(3x2 + 5x − 7)(x − 1)

By column method:

(x2 − 2x + 3)(x + 4)

By column method:

(3x2 + 5x − 7)(x − 1) − (x2 − 2x + 3)(x + 4)

View NCERT Solutions for all chapters of Class 7